重载类型扣除的赋值运算符

问题描述

以下是ideone程式码：

我的问题是，是否可以强制基于左值单独的转换？例如，

  [秒] s = 2_h + 60_s; 
 cout<< s.getSeconds（）<< endl; 
  

显然，我必须写一些类似2_h.toSeconds

解决方案

要允许这样做（这比你写的更可能是你的问题，如果我错了）：

 秒s = 2_h; 
  

以下操作将会起作用：Add operator Seconds（）const 到类小时：

  b unsigned long long_hours; 
 public：
小时（unsigned long long hours）：_hours（hours）{} 
 
操作符Seconds（）const; 
 
无符号长整型getHours（）const {
 return this-> _hours; 
} 
}; class  Seconds  之后定义   code>：
  Hours :: operator Seconds（）const {return this-> _hours * 3600; } 
  
 
Here's the ideone code: http://ideone.com/Qp8Eqg
My question is, is it possible to force a conversion based on the lvalue alone? For example, 
[Seconds] s = 2_h + 60_s;
cout <<s.getSeconds()<<endl;
Obviously, I would have to write something like 2_h.toSeconds(), but that would be too verbose and doesn't achieve the idea.
 解决方案 
To allow this (which is more likely your question than what you wrote, correct me if I'm wrong):
Seconds s = 2_h;
the following would work: Add operator Seconds() const to class Hours:
class Hours {
    unsigned long long _hours;
public:
    Hours(unsigned long long hours) : _hours(hours) { }

    operator Seconds() const;

    unsigned long long getHours() const {
        return this->_hours;
    }
};
and define it after class Seconds:
Hours::operator Seconds() const { return this->_hours * 3600; }
                        
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